Question

Find the highest power of 5 dividing 100! and find number of zero at the end of 100!

Answer 1

Highest exponent of a prime number a in n! is

$=\left[\frac{n}{a}\right]+\left[\frac{n}{a^2}\right]+\left[\frac{n}{a^3}\right]+....$

where [.] is greatest integer formation

$\Rightarrow$ Exponent of $5$ in $100!$

is $=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]+\left[\frac{100}{5^3}\right]+....$

$=20+4+0+0....$

$=24$

To make a zero in the end, use need a 10 as a factor

i.e., a product of $2$ and $5$

Exponent of 2 in $100!$ will be more than that of $5$, as $2$ is smaller number.

Hence $5^{24}$ will multiply with $2^{24}$ to make ${10}^{24}$

$\Rightarrow 24$ zeroes are there at the end of $100!$